$ g(x) = \int_{-\pi}^{x}\sin(t)\,dt\,$ $ g\,'(\pi)\, =$
The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \sin(t) $ is continuous on $[-\pi,\pi]$. Applying the theorem We're given: $ g(x) = \int_{-\pi}^{x}\sin(t)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =\sin(x)$ Evaluating $g'(\pi)$ $ g'(\pi)= \sin(\pi) = 0$ The answer: $g'(\pi)=0$